Distribution identity

For $a>0$ and $t>0$,

$$ \mathbb P\left(\max_{0\le s\le t} B_s\ge a\right)

\mathbb P(|B_t|\ge a). $$

Interpretation

The running maximum of Brownian motion over $[0,t]$ has the same distribution as the absolute value of the endpoint $|B_t|$.

Source

This follows immediately from the reflection principle:

$$ \mathbb P\left(\max_{0\le s\le t} B_s\ge a\right)

\mathbb P(T_a\le t)

2\mathbb P(B_t\ge a)

\mathbb P(|B_t|\ge a). $$